Compute the following limits using L'Hopital's Rule if appropriate.
$$\lim_{x \rightarrow \infty} \frac{ \ln(x^5-9) }{ \ln(x) }$$
and
$$\lim_{x \rightarrow \infty}\frac{ e ^{9x} }{ e ^{10x}-e ^{-10x} }$$

Question

Compute the following limits using L'Hopital's Rule if appropriate.
$$\lim_{x \rightarrow \infty} \frac{ \ln(x^5-9) }{ \ln(x) }$$ 
and
$$\lim_{x \rightarrow \infty}\frac{ e ^{9x} }{ e ^{10x}-e ^{-10x} }$$

anonymous · Answer

How do you figure out if the limit is infinity/infinity?

anonymous · Answer

@hartnn

hartnn · Answer

plug in the value of x,

 $\ln \infty = \infty \ e^\infty = \infty $

hartnn · Answer

so both the limits are indeed of the form infinity/infinity, 
L'Hopital Rule can be used.

anonymous · Answer

(ln(x^(5)-9))' = 5x^4/x^5-9
(ln(x))'=1/x

hartnn · Answer

yes.

now check whether this is on the form of infinity/infinity or 0/0 

note: 1/ infinity = 0

anonymous · Answer

So the 5x^4/(x^-9) is going to infinity and 1/x is 0

anonymous · Answer

So the final answer is 0??

hartnn · Answer

the numerator is 5x^4/(x^5-9) is equivalent to 5x^4/x^5 for large values of x,

so, 5/x or 5/infinity = 0 

so, the limit is indeed of the form 0/0 
makes sense?

anonymous · Answer

Okay.. Kind of get it